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Concept 02 / 06Vector Algebra

Vector (Cross) Product: Area, Direction and Properties

The definition |a x b| = |a||b| sin theta with the right-hand rule, the determinant formula, anti-commutativity, parallelism test, area of parallelograms and triangles, unit vectors perpendicular to two vectors, moment of a force, and Lagrange's identity.

Figure 1
The vector product of a⃗\vec{a} and b⃗\vec{b} is the vector a⃗×b⃗=∣a⃗∣∣b⃗∣sin⁡θ n^\vec{a} \times \vec{b} = |\vec{a}||\vec{b}|\sin\theta\,\hat{n}, where θ∈[0,π]\theta \in [0, \pi] is the angle between them and n^\hat{n} is the unit vector perpendicular to both, chosen by the right-hand rule: curl the fingers of the right hand from a⃗\vec{a} towards b⃗\vec{b} through the smaller angle, and the thumb points along n^\hat{n}. Its magnitude equals the area of the parallelogram spanned by a⃗\vec{a} and b⃗\vec{b}, as in the figure. Component (determinant) formula: a⃗×b⃗=∣i^j^k^a1a2a3b1b2b3∣=(a2b3−a3b2)i^−(a1b3−a3b1)j^+(a1b2−a2b1)k^\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{vmatrix} = (a_{2}b_{3} - a_{3}b_{2})\hat{i} - (a_{1}b_{3} - a_{3}b_{1})\hat{j} + (a_{1}b_{2} - a_{2}b_{1})\hat{k}. It follows from i^×j^=k^\hat{i} \times \hat{j} = \hat{k}, j^×k^=i^\hat{j} \times \hat{k} = \hat{i}, k^×i^=j^\hat{k} \times \hat{i} = \hat{j} (cyclic order), the reverse orders giving negatives, and i^×i^=0⃗\hat{i} \times \hat{i} = \vec{0} and so on. Properties. Anti-commutative: b⃗×a⃗=−(a⃗×b⃗)\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b}). The order matters. Distributive: a⃗×(b⃗+c⃗)=a⃗×b⃗+a⃗×c⃗\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}. Not associative: a⃗×(b⃗×c⃗)≠(a⃗×b⃗)×c⃗\vec{a} \times (\vec{b} \times \vec{c}) \neq (\vec{a} \times \vec{b}) \times \vec{c} in general. For instance i^×(i^×j^)=i^×k^=−j^\hat{i} \times (\hat{i} \times \hat{j}) = \hat{i} \times \hat{k} = -\hat{j}, while (i^×i^)×j^=0⃗(\hat{i} \times \hat{i}) \times \hat{j} = \vec{0}. a⃗×a⃗=0⃗\vec{a} \times \vec{a} = \vec{0}, and for non-zero vectors a⃗×b⃗=0⃗  ⟺  a⃗∥b⃗\vec{a} \times \vec{b} = \vec{0} \iff \vec{a} \parallel \vec{b}. This is the standard parallelism test. a⃗×b⃗\vec{a} \times \vec{b} is perpendicular to both a⃗\vec{a} and b⃗\vec{b}, so (a⃗×b⃗)⋅a⃗=(a⃗×b⃗)⋅b⃗=0(\vec{a} \times \vec{b}) \cdot \vec{a} = (\vec{a} \times \vec{b}) \cdot \vec{b} = 0. No cancellation: a⃗×b⃗=a⃗×c⃗\vec{a} \times \vec{b} = \vec{a} \times \vec{c} only means a⃗∥(b⃗−c⃗)\vec{a} \parallel (\vec{b} - \vec{c}). Lagrange's identity: ∣a⃗×b⃗∣2+(a⃗⋅b⃗)2=∣a⃗∣2∣b⃗∣2|\vec{a} \times \vec{b}|^{2} + (\vec{a} \cdot \vec{b})^{2} = |\vec{a}|^{2}|\vec{b}|^{2}, which is sin⁡2θ+cos⁡2θ=1\sin^{2}\theta + \cos^{2}\theta = 1 scaled. It gives ∣a⃗×b⃗∣|\vec{a} \times \vec{b}| without computing the cross product. Applications. Area of the parallelogram with adjacent sides a⃗,b⃗\vec{a}, \vec{b}: ∣a⃗×b⃗∣|\vec{a} \times \vec{b}|. With diagonals d⃗1,d⃗2\vec{d}_{1}, \vec{d}_{2}: 12∣d⃗1×d⃗2∣\frac{1}{2}|\vec{d}_{1} \times \vec{d}_{2}|. Area of triangle ABCABC: 12∣AB⃗×AC⃗∣=12∣a⃗×b⃗+b⃗×c⃗+c⃗×a⃗∣\frac{1}{2}|\vec{AB} \times \vec{AC}| = \frac{1}{2}|\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}| in terms of position vectors. Unit vectors perpendicular to both a⃗\vec{a} and b⃗\vec{b}: ±a⃗×b⃗∣a⃗×b⃗∣\pm\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} (two, opposite). sin⁡θ=∣a⃗×b⃗∣∣a⃗∣∣b⃗∣\sin\theta = \frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}. Moment (torque) of a force F⃗\vec{F} acting at a point with position vector r⃗\vec{r} about the origin: M⃗=r⃗×F⃗\vec{M} = \vec{r} \times \vec{F}. Angular momentum L⃗=r⃗×p⃗\vec{L} = \vec{r} \times \vec{p} and the velocity of a point in a rotating body v⃗=ω⃗×r⃗\vec{v} = \vec{\omega} \times \vec{r} have the same structure. Key Takeaways: - ∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta = parallelogram area; direction by the right-hand rule. - Determinant formula with i^,j^,k^\hat{i}, \hat{j}, \hat{k} in the first row. - b⃗×a⃗=−a⃗×b⃗\vec{b} \times \vec{a} = -\vec{a} \times \vec{b}; not associative. - a⃗×b⃗=0⃗  ⟺  a⃗∥b⃗\vec{a} \times \vec{b} = \vec{0} \iff \vec{a} \parallel \vec{b} (non-zero vectors). - ∣a⃗×b⃗∣2+(a⃗⋅b⃗)2=∣a⃗∣2∣b⃗∣2|\vec{a} \times \vec{b}|^{2} + (\vec{a} \cdot \vec{b})^{2} = |\vec{a}|^{2}|\vec{b}|^{2}. - Triangle area 12∣AB⃗×AC⃗∣\frac{1}{2}|\vec{AB} \times \vec{AC}|; perpendicular unit vectors ±a⃗×b⃗∣a⃗×b⃗∣\pm\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}. - Pitfall: sign errors in the middle (j^\hat{j}) term of the determinant. Brain-Teaser 1: Find a unit vector perpendicular to both a⃗=i^+2j^−k^\vec{a} = \hat{i} + 2\hat{j} - \hat{k} and b⃗=2i^−j^+k^\vec{b} = 2\hat{i} - \hat{j} + \hat{k}, and the area of the triangle they span. Solution: a⃗×b⃗=(2⋅1−(−1)(−1))i^−(1⋅1−(−1)⋅2)j^+(1⋅(−1)−2⋅2)k^=i^−3j^−5k^\vec{a} \times \vec{b} = (2 \cdot 1 - (-1)(-1))\hat{i} - (1 \cdot 1 - (-1) \cdot 2)\hat{j} + (1 \cdot (-1) - 2 \cdot 2)\hat{k} = \hat{i} - 3\hat{j} - 5\hat{k}. Its magnitude is 1+9+25=35\sqrt{1 + 9 + 25} = \sqrt{35}. Unit vectors: ±i^−3j^−5k^35\pm\frac{\hat{i} - 3\hat{j} - 5\hat{k}}{\sqrt{35}}. Triangle area: 352\frac{\sqrt{35}}{2}. Check perpendicularity: (i^−3j^−5k^)⋅a⃗=1−6+5=0(\hat{i} - 3\hat{j} - 5\hat{k}) \cdot \vec{a} = 1 - 6 + 5 = 0. Brain-Teaser 2: If ∣a⃗∣=2|\vec{a}| = 2, ∣b⃗∣=5|\vec{b}| = 5 and ∣a⃗×b⃗∣=8|\vec{a} \times \vec{b}| = 8, find a⃗⋅b⃗\vec{a} \cdot \vec{b}. Solution: Lagrange's identity: (a⃗⋅b⃗)2=4×25−64=36(\vec{a} \cdot \vec{b})^{2} = 4 \times 25 - 64 = 36, so a⃗⋅b⃗=±6\vec{a} \cdot \vec{b} = \pm 6. The sign cannot be decided from the data: the angle could be θ\theta or π−θ\pi - \theta with the same sine. Both answers are correct unless the angle is stated to be acute or obtuse. Brain-Teaser 3: If a⃗×b⃗=c⃗×d⃗\vec{a} \times \vec{b} = \vec{c} \times \vec{d} and a⃗×c⃗=b⃗×d⃗\vec{a} \times \vec{c} = \vec{b} \times \vec{d}, show that a⃗−d⃗\vec{a} - \vec{d} is parallel to b⃗−c⃗\vec{b} - \vec{c}. Solution: (a⃗−d⃗)×(b⃗−c⃗)=a⃗×b⃗−a⃗×c⃗−d⃗×b⃗+d⃗×c⃗(\vec{a} - \vec{d}) \times (\vec{b} - \vec{c}) = \vec{a} \times \vec{b} - \vec{a} \times \vec{c} - \vec{d} \times \vec{b} + \vec{d} \times \vec{c}. Using d⃗×b⃗=−b⃗×d⃗\vec{d} \times \vec{b} = -\vec{b} \times \vec{d} and d⃗×c⃗=−c⃗×d⃗\vec{d} \times \vec{c} = -\vec{c} \times \vec{d}, this is (a⃗×b⃗−c⃗×d⃗)−(a⃗×c⃗−b⃗×d⃗)=0⃗−0⃗=0⃗(\vec{a} \times \vec{b} - \vec{c} \times \vec{d}) - (\vec{a} \times \vec{c} - \vec{b} \times \vec{d}) = \vec{0} - \vec{0} = \vec{0}. A zero cross product means the vectors are parallel (or one is zero). Expanding carefully with anti-commutativity is the whole proof.

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